Find the line of reflection that maps the trapezoid, formed with vertices at #A(-3,1)#, #B(-1,1)#, #C(0,4)# and #D(-4,4)#, onto itself?

1 Answer
Dec 19, 2016

#x+2=0#

Explanation:

The trapezoid is formed with its vertices at #A(-3,1)#, #B(-1,1)#, #C(0,4)# and #D(-4,4)#. These have been named to add clarity to answer.

As ordinates of #A# and #B# are same, as also ordinates of #C#and #D#, it the lines #AB# and #CD# are parallel to #x#-axiss and hence it is apparent that #AB#||#CD#.

Further #AD=sqrt((-3-(-4))^2+(1-4)^2)=sqrt10# and
#BC=sqrt((-1-0)^2+(1-4)^2)=sqrt10# and as such #AD=BC# and it is an isosceles trapezium.

Hence the line joining the midpoints of #AB# and #CD# i.e.#(-2,1)# and #(-2,4)# is the line of reflection that maps the trapezoid onto itself, which is parallel to #y#-axis.

and its equation is #x=-2# or #x+2=0#.